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Yahoo!

  • 3 years ago

The x co-ordinate of a point on the curve 9y^2 = x^3 , the normal at which cuts off equal intercept on the axes is ?

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  1. sauravshakya
    • 3 years ago
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    9y^2=x^3 d(9y^2)/dx =d(x^3)/dx 18y dy/dx =3x^2 dy/dx =x^2/(6y)

  2. sauravshakya
    • 3 years ago
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    The gradient of tangent is x^2/(6y) So, gradient of normal is -6y/(x^2)

  3. sauravshakya
    • 3 years ago
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    NOW, let x-intercept of the normal be a then y-intercept is also a.

  4. sauravshakya
    • 3 years ago
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    Gradient of normal = (a-0)/(0-a) -6y/(x^2) =-1

  5. sauravshakya
    • 3 years ago
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    Now,|dw:1359201825424:dw| and solve for x

  6. sauravshakya
    • 3 years ago
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    |dw:1359201971611:dw|

  7. sauravshakya
    • 3 years ago
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    U need to solve that

  8. Yahoo!
    • 3 years ago
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    |dw:1359202095063:dw|

  9. sauravshakya
    • 3 years ago
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    9y^2=x^3

  10. Yahoo!
    • 3 years ago
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    -6y/(x^2) =-1......then wat did u do ?

  11. sauravshakya
    • 3 years ago
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    Substitute y interms of x

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